leetcode/apps/docs/content/binary-search-tree/98-validate-binary-search-tree.mdx

---
title: '98. Validate Binary Search Tree'
description: Given the root of a binary tree, determine if it is a valid binary search tree (BST)
sidebar:
  label: 'Validate Binary Search Tree'
  badge: 'Medium'
---

<Badge variant="accent">Tree</Badge>

### Example 1:
- Input: `root = [2,1,3]`
- Output: `true`

### Example 2:
- Input: `root = [5,1,4,null,null,3,6]`
- Output: `false`
- Explanation: The `root` node's value is `5` but its right child's value is `4`.

### Constraints:

- The number of nodes in the tree is in the range [1, 10^4].
- `-2^31 <= Node.val <= 2^31 - 1`

## Approach

```mermaid
flowchart TD
  S(["isValidBST(root)"]) --> C["valid(root, float('-inf'), float('inf')) — the root may hold any value"]
  C --> E(["return that result"])
  subgraph valid["valid(node, low, high)"]
    V1{"not node?"}
    V1 -- yes --> V2(["return True"])
    V1 -- no --> V3{"not (low < node.val < high)?"}
    V3 -- yes --> V4(["return False — node breaks a bound set by an ancestor"])
    V3 -- no --> V5["valid(node.left, low, node.val) and valid(node.right, node.val, high) — each child tightens one bound"]
    V5 --> V6(["return that result"])
  end
  C --> V1
```

## Solution

```py
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isValidBST(self, root: Optional[TreeNode]) -> bool:
        def valid(node, low, high):
            if not node:
                return True

            if not (low < node.val < high):
                return False

            return valid(node.left, low, node.val) and valid(node.right, node.val, high)

        return valid(root, float("-inf"), float("inf"))
```

## Explanation

<YouTube id="s6ATEkipzow" title="Validate Binary Search Tree - Depth First Search - Leetcode 98" />